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How do you differentiate #g(y) =(2x^3+x )e^x # using the product rule?

Answer 1

Elijah Abram

If #g(y)#: #g'(y) = 0#

If #g(x)#: #g'(x) = (2x^3+7x+1) e^x#

#g(y) = (2x^3 +x)e^x#

#g'(y) = 0#

If you meant #g(x) = (2x^3+x)e^x#, then

#g'(x) = d/(dx)[(2x^3+x)]e^x + (2x^3+x)d/(dx)[e^x]#

#g'(x) = (6x+1)e^x + (2x^3+x)e^x#

#g'(x) = (6x+1 + 2x^3 + x)e^x#

#g'(x) = (2x^3+7x+1) e^x#

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Answer 2

Samantha Adkison

To differentiate ( g(y) = (2x^3 + x)e^x ) using the product rule:

Identify the two functions being multiplied: ( f(x) = 2x^3 + x ) and ( h(x) = e^x ).
Apply the product rule: ( g'(y) = f'(x)h(x) + f(x)h'(x) ).
Differentiate ( f(x) ) with respect to ( x ) to get ( f'(x) ).
Differentiate ( h(x) ) with respect to ( x ) to get ( h'(x) ).
Substitute the derivatives and the original functions into the product rule formula.
Simplify the expression to get the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.